Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}3x+3y &= -8 \\ -9x+y &= 4\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-9x = -y+4$ Divide both sides by $-9$ to isolate $x$ $x = {\dfrac{1}{9}y - \dfrac{4}{9}}$ Substitute this expression for $x$ in the first equation. $3({\dfrac{1}{9}y - \dfrac{4}{9}}) + 3y = -8$ $\dfrac{1}{3}y - \dfrac{4}{3} + 3y = -8$ Simplify by combining terms, then solve for $y$ $\dfrac{10}{3}y - \dfrac{4}{3} = -8$ $\dfrac{10}{3}y = -\dfrac{20}{3}$ $y = -2$ Substitute $-2$ for $y$ in the top equation. $3x+3( -2) = -8$ $3x-6 = -8$ $3x = -2$ $x = -\dfrac{2}{3}$ The solution is $\enspace x = -\dfrac{2}{3}, \enspace y = -2$.